POJ2586Y2K Accounting Bug(贪心 + 不好想)

Y2K Accounting Bug

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12251		Accepted: 6202

Description

Accounting for Computer Machinists (ACM) has sufferred from the Y2K bug and lost some vital data for preparing annual report for MS Inc.
All what they remember is that MS Inc. posted a surplus or a deficit each month of 1999 and each month when MS Inc. posted surplus, the amount of surplus was s and each month when MS Inc. posted deficit, the deficit was d. They do not remember which or how many months posted surplus or deficit. MS Inc., unlike other companies, posts their earnings for each consecutive 5 months during a year. ACM knows that each of these 8 postings reported a deficit but they do not know how much. The chief accountant is almost sure that MS Inc. was about to post surplus for the entire year of 1999. Almost but not quite.

Write a program, which decides whether MS Inc. suffered a deficit during 1999, or if a surplus for 1999 was possible, what is the maximum amount of surplus that they can post.

Input

Input is a sequence of lines, each containing two positive integers s and d.

Output

For each line of input, output one line containing either a single integer giving the amount of surplus for the entire year, or output Deficit if it is impossible.

Sample Input

59 237
375 743
200000 849694
2500000 8000000

Sample Output

116
28
300612
Deficit
题意：12个月每个月要么盈利s，要么亏损d，连续五个月都是亏损的，即1-5，2-6,3-7,4-8,5-9,6-10,7-11,8-12；求一年最大盈利
分析：想了半天确实没想出怎么解，看了题解，却又感觉那么的合情合理，贪心考虑边界http://blog.csdn.net/lyy289065406/article/details/6642603

在保证连续5个月都亏损的前提下，使得每5个月中亏损的月数最少。

              x=1:  ssssd,ssssd,ss    d>4s     赢利10个月  10s-2d
              x=2:  sssdd,sssdd,ss    2d>3s    赢利8个月     8s-4d
              x=3:  ssddd,ssddd,ss    3d>2s    赢利6个月     6s-6d 
              x=4:  sdddd,sdddd,sd    4d>s     赢利3个月     3s-9d
              x=5:  ddddd,ddddd,dd    4d<s     无赢利

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
long long s,d;
char str[10] = "Deficit";

int main()
{
    while(scanf("%I64d%I64d", &s, &d) != EOF)
    {
        if(d > 4 * s)
        {
            if(10 * s - 2 * d > 0)
            {
                printf("%I64d\n", 10 * s - 2 * d);
            }
            else
            {
                printf("%s\n", str);
            }
        }
        else if(3 * s < 2 * d )
        {
            if(8 * s - 4 * d > 0)
            {
                printf("%I64d\n", 8 * s - 4 * d);
            }
            else
            {
                printf("%s\n", str);
            }
        }
        else if(2 * s < 3 * d)
        {
            if(6 * s - 6 * d > 0)
            {
                printf("%I64d\n", 6 * s - 6 * d);
            }
            else
            {
                printf("%s\n", str);
            }
        }
        else if(s < 4 * d)
        {
            if(3 * s - 9 * d > 0)
            {
                printf("%I64d\n", 3 * s - 9 * d);
            }
            else
            {
                printf("%s\n", str);
            }
        }
        else
        {
            printf("%s\n", str);
        }
    }

    return 0;
}